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I have a Fibonacci problem ... basically I have to show that:

$$F(n+2) = 1 + \sum_{i=0}^{n} F(i)$$

Using strong induction, I must now show that:

$$F(n+2) = F(n+1) + F(n)$$ $$F(n+2) = (1 + \sum_{i=0}^{n-1} F(i)) + (1 + \sum_{i=0}^{n-2} F(i))$$ $$F(n+2) = 2 + \sum_{i=0}^{n-1} F(i) + \sum_{i=0}^{n-2} F(i)$$

How do I add these two summations to show that they equal the first equation (at the top)?

Please provide the steps with explanation as to what you are doing. Thanks

J. M. ain't a mathematician
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K Split X
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2 Answers2

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You’re making it much harder than it really is. In particular, you don’t need strong induction. Your induction hypothesis is that

$$F(n+2)=1+\sum_{k=0}^nF(k)\;.$$

Now add $F(n+1)$ to both sides and simplify each side:

$$F(n+3)=F(n+2)+F(n+1)=1+\sum_{k=0}^nF(k)+F(n+1)=1+\sum_{k=0}^{n+1}F(k)\;.$$

The equality of the leftmost and rightmost quantities in that string is exactly what you needed to prove for your induction step.

Brian M. Scott
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    In this case, you are using simple induction, and going from n --> n+1? – K Split X Dec 11 '16 at 02:24
  • @KracX: Yes, exactly. There’s no reason to invoke so-called strong induction here. – Brian M. Scott Dec 11 '16 at 02:25
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    We did it using strong induction in class, that's why I asked that way, but I guess simple induction is way easier – K Split X Dec 11 '16 at 02:26
  • I'm not sure who voted you down, but I gave you a vote up right now – K Split X Dec 11 '16 at 02:27
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    @KracX: It’s okay; in fact the downvote was removed and might even have been an accident. \ In general you shouldn’t worry about just what kind of induction you’re using. You figure out how many of the earlier stages you need in order to make the induction step work, and you use that many. If you need only one, you have ordinary induction. If you need them all, you have strong induction. And quite often you have some fixed number greater than one — two, quite often, if you’re working with Fibonacci numbers. – Brian M. Scott Dec 11 '16 at 02:27
  • Yeah because Fibonacci is recursive, and in recursive, almost $always$ we need to use strong induction so we can assume base cases before – K Split X Dec 11 '16 at 02:44
  • Adding onto your point, "And quite often you have some fixed number greater than one — two, quite often, if you’re working with Fibonacci numbers" ... this is why I did strong – K Split X Dec 11 '16 at 02:45
  • @KracX: But my point is that when you reach back just two steps every time, what you’re doing is more like ordinary induction than it is like a true strong induction in which you need all earlier cases. Books and teachers do students a disservice when they insist on emphasizing the distinction between ordinary and strong induction. It’s really all the same thing, just with slightly different details. – Brian M. Scott Dec 11 '16 at 02:46
  • @KracX if you satisfied with the answer given you have to accept it not just up vote – IrbidMath Dec 11 '16 at 03:55
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If you are asking for any induction proof for this identity, then your question is a duplicate of some already existing questions, for example, Fibonacci using proof by induction: $\sum_{i=1}^{n-2}F_i=F_n-2$ (and other posts linked there) or For the Fibonacci sequence prove that $\sum_{i=1}^n F_i= F_{n+2} - 1$ (and other posts linked there

If you are asking specifically whether the approach you've chosen can be somehow finished, the answer is that it is possible.

From the result you are trying to prove it seems that you are using $F(0)=0$ and $F(1)=1$. (You should probably specify this in your question, since $F(0)=F(1)=1$ is quite common too.)

You have \begin{align*} F(n+2) &\overset{(1)}= 2 + \sum_{i=0}^{n-1} F(i) + \sum_{i=0}^{n-2} F(i)\\ &\overset{(2)}= 2 + \sum_{i=1}^{n} F(i-1) + \sum_{i=2}^{n} F(i-2)\\ &\overset{(3)}= 1+ F(-1) +\sum_{i=1}^{n} F(i-1) + F(-2)+ F(-1) +\sum_{i=2}^{n} F(i-2)\\ &\overset{(4)}= 1 + \sum_{i=0}^{n} F(i-1) + \sum_{i=0}^{n} F(i-2)\\ &\overset{(5)}= 1 + \sum_{i=0}^{n} (F(i-1) + F(i-2))\\ &\overset{(6)}= 1 + \sum_{i=0}^{n} F(i) \end{align*}

Notice that if you are using $F(0)=F(1)=1$ (as you mentioned in a comment under this answer - now deleted), then you get $F(-1)=F(1)-F(0)=0$ and similarly $F(-2)=F(0)-F(-1)=1$.

I will add that that the same approach would work with $F(0)=0$ and $F(1)=1$, which is probably more usual. (This are the initial values mentioned in the Wikipedia article about Fibonacci numbers.) In this case we would get $F(-2)=-1$, $F(-1)=1$ and again $2=1+F(-1)+F(-2)+F(-1)$. With these initial values we have $F(-n)=(-1)^{n+1}F(n).)

$(3)$: Here we simply used $2=1+F(-1)=1+F(-1)+F(-2)+F(-1)$. (Which is true since $F(-1)=0$ and $F(-2)=1$.) We are doing this because these are precisely the numbers missing in the sums to get a sum from $0$ to $n$.

After posting this answer I found a rather similar approach in this answer: Fibonacci proof question $\sum_{i=1}^nF_i = F_{n+2} - 1$. You might have a look at the way it is written there to see whether it could make things clearer.

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Martin Sleziak
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